## Pseudo-first-order reaction

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Catherine Liu 1D
Posts: 9
Joined: Mon Oct 03, 2016 3:00 am

### Pseudo-first-order reaction

What exactly is a pseudo-first-order reaction?

Sunny Chera 1N
Posts: 20
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Pseudo-first-order reaction

When rate laws depend on more than one species, we can simplify the analysis by keeping on the species at a "constant" concentration throughout the reaction. For example, using the textbook's example, iodide ions get oxidized by persulfate ions, represented by: Rate of consumption of I^-=k[S2O8^2-][I^-]. The concentration of S2O8^2- can be raised so high that in the course of the reaction, their concentration change will be negligible. Let's say the concentration of these persulfate ions was a 100x the concentration of the iodide ions. Then, even when the iodide ions have been all used up (oxidized), the concentration of the persulfate ions would be pretty much the same. Not exactly, but pretty much. Therefore, we can exclude the concentration of the persulfate ions from the rate law. As a result, the Rate of consumption of I^-=k' [I^-]. k' would be k[S2O8^2-], the concentration that is now excluded. Now the rate of consumption of I^- purely depends on the concentration of the iodide ions, which is A LOT easier to analyze than if both concentrations were taken into consideration, as the second-order reaction is turned into a "pseudo-first-order reaction" by removing the concentration of the persulfate ions. To answer your question in full, a pseudo-first-order reaction results from a second-order reaction that has one of its species' concentration kept constant.